5^2 and 3^3 are factors of n* 2^5 * 6^2 * 7^3
Prime factorize 6^2 = 2^2 * 3^2
n*2^5 * 6^2 * 7^3 => n*2^5 * 2^2 * 3^2 * 7^3
for 5^2 and 3^3 to be factor of above we need atleast two 5's and three 3's
we already have two 3's but no 5's
Therefore n = 5^2 * 3 = 25*3 = 75
Hope this helps.
smallest possible value
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parallel_chase
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n*(2^5)*(6^2)*(7^3) might be written in the following manner: n*(2^7)*(3^2)*(7^3). The part of this number (2^7)*(3^2)*(7^3) does not have any "5" and one additional "3" that are necessary to satisfy the statement. It means that "n" has two "5" and one "3". So, the smallest value of n is (3^2)*5.
